Problem: $\begin{aligned} &f(x)=3\cdot2^x \\\\ &h(x)=2x-7 \end{aligned}$ $h(f(2))=$
Solution: When evaluating composite functions, we work our way inside out. To evaluate $h(f(2))$, let's first evaluate $f(2)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $f({2})$. $\begin{aligned}f(x)&=3\cdot 2^x\\\\ f({2})&=3\cdot 2^{{2}} ~~~~~~~~~~\text{Plug in }x={2}\\\\ &=3\cdot 4\\\\ &={12}\end{aligned}$ We now know that $h(f({2}))$ is the same as $h({12})$ because $f({2}) = {12}$. Let's evaluate $h({12})$. $\begin{aligned}h(x)&=2x-7\\\\ h({{12}})&=2({12})- 7 ~~~~~~~~~~\text{Plug in }x={12}\\\\ &=24 - 7\\\\ &=17\end{aligned}$ The answer: $h(f(2)) = 17$